← Group Work

Problem 1

1. I get $\fbox{$\beta=85\,\mbox{dB}$}$.
2. I get $\fbox{$(\Delta t)=0.15\,\mbox{s}$}$, and (quite likely) "yes".

Problem 2

1. I get $\fbox{$L=3.0\,\mbox{m}$}$.

Problem 3

1. See the formula above. When the Bungee cord is stretched, the tension increases AND the linear mass density (how many kilograms per meter) decreases. The one that is stretched more (second one) should have the higher velocity.
2. I get the ratio of fundamental frequencies to be $\fbox{$f_1/f_2 = 1/\sqrt{1.5} = 0.82 $}$. The one that is stretched less has the lower fundamental frequency.

Problem 1

1. I get $\fbox{$v_{\mbox{max}}=28\,\mbox{m/s}$}$. [Where does this maximum speed occur in the ride?]
2. Sketch the diagram.
3. I get $\fbox{$(5T)=62\,\mbox{sec}$}$.

Hints <SHOW>

Problem 2

Problem 3

1. I get $\fbox{$\lambda=1.14\,\mbox{m}$}$.
2. I get $\fbox{$v=342\,\mbox{m/s}$}$.
3. I get $\fbox{$\lambda=0.57\,\mbox{m}$}$.
4. I get $\fbox{$\lambda=5.10\,\mbox{m}$}$.

Problem 1

1. I get $\fbox{$(T/4=6.1\,\mbox{s}$}$. (Where is the $8^\circ$ part used in this problem?)
2. I get $\fbox{$(T/4=6.1\,\mbox{s}$}$. (Same as previous since it is still a small angle.)

Problem 2

1. I get $\fbox{$L{_\mbox{(Ohio)}}=0.9929\,\mbox{m}$}$.
2. I get $\fbox{$L{_\mbox{(Peru)}}=0.9893\,\mbox{m}$}$.
3. When both clocks are in Ohio, the shorter one (adjusted in Peru) will run faster since it has a shorter period ($T=2\pi\sqrt{L/g}$).

Problem 1

1. velocities ranked: fastest 2 > 3 > 1 slowest
2. diameters ranked: largest 2 > 1 > 3 smallest

Problem 2

Problem 3

Here is a picture of paper lanterns I got somewhere from the internet

Problem 1

1. $\fbox{$N_1=0\,\mbox{N}$}$ (on verge of tipping)
   $\fbox{$N_2=900\,\mbox{N}$}$ (supporting all weight above it, and 2 sig-figs $9.0\times 10^2\,\mbox{N}$)
2. $\fbox{$s=0.041\,\mbox{m}$}$

Problem 2

Problem 3

Problem 1

1. The static equilibrium formula for torque should be \[ (\Sigma\tau)_{\mathcal{O}} = (2.0)\cdot(T\,\sin(40^\circ))-((15)(9.8))(1.0)-((20)(9.8))(1.8) = 0 \]

   where axis $\mathcal{0}$ is the left end of the beam. This leads to the value for the tension to be $\fbox{$T=389\,\mbox{N}$}$.

2. The strategy is to use $(\Sigma F)_x=0$ and $(\Sigma F)_y=0$ to find the two components of the unknown force that the wall exerts on the beam. We can actually do it to get: \begin{align} & (\Sigma F)_x = F_x - T \,\cos(40^\circ)=0 \qquad\rightarrow\qquad F_x = 298\,\mbox{N} \\[2ex] &(\Sigma F)_y = F_y + T\,\sin(40^\circ) - ((15)(9.8)) - ((20)(9.8)) = 0 \qquad\rightarrow\qquad F_y = 93\,\mbox{N} \end{align}

   Here is a to-scale free body digram for the beam:

   

Problem 1

1. I get a relatively simple formula that looks like: \[ v = \sqrt{\frac{2g\,r\,\Delta \theta}{1+\tfrac12 M_p/m}} \]

   It might be easier and quicker to solve it numerically in steps.

   This gives a speed of $\fbox{$v=1.3\,\mbox{m/s}$}$.

   (When you study this problem later, you can compare it to $v=\sqrt{2gh}$ which we get for the speed of an object that is dropped through a distance $h$.)

2. Yes. It is possible to get the answer in a similar number of steps, or more steps. (What are those other physics prinicples?)

Problem 2

Problem 3

1. I get $\fbox{$\omega_f=0.71\,\mbox{rad}/\mbox{s}$}$, the playground toy slows down.
2. I get $\fbox{$\Sigma\tau = -41\,mbox{N}\,\mbox{m}$}$.

Problem 1

1. Use $(\Sigma\tau)=I\alpha$: \begin{align} & I = \tfrac25\,M R^2 = \tfrac25\,(1800)(0.50)^2 = 180\,\mbox{kg}\,\mbox{m}^2 \\[1ex] & (\Delta\theta)=\omega_0\cdot(\Delta t)+\tfrac12\,\alpha\,\cdot(\Delta t)^2 \qquad \rightarrow \qquad \alpha = \frac{(\pi)}{(144)} = 0.0218\,\mbox{s}^{-2} \\[1ex] & (\Sigma\tau) = (180)(0.0218) = 3.92\,\mbox{N}\,\mbox{m} \end{align}
2. Use $(\Sigma\tau) = (\Sigma F)\cdot R$ to get \[ (\Sigma F) = \frac{(3.92)}{(0.50)} = 7.85\,\mbox{N} \]
3. The weight of an apple (the fruit, or the phone brand) is somewhere between one and two Newtons. (My Android phone is 230 grams which is about $2.3\,\mbox{N}$. Comparing the force above to a one-Newton apple, it is almost eight times bigger than that. Is that reasonable?
4. Zero. Explain why. The weight of the stone is about 18,000 Newtons and its radius is half of one meter. How can the torque be zero? (Being pedantic: The torque would be caused by the Earth via the gravitational force, a.k.a. the stone's weight; torque is not exerted by gravity.)

Problem 1

1. $\fbox{$t=62\,\mbox{s}$}$
2. It will affect the result, but the question becomes, with more coins tossed, and less mass remaining, will it move faster or slower? We can calculate it in the next part.
3. $\fbox{$t=41\,\mbox{s}$}$

Going a little further <SHOW>

Chaning the problem <SHOW>

Problem 2

If you have trouble, a further hint is that you can validate the numbers using \[ (75)(15) = (67)(10) + (15)(30) \] with a two-digit precision. In that, I assume that the pivot point is at the string attached to the ceiling above.

Problem 3

1. $\fbox{$\alpha=1130\,\mbox{s}^{-2}$}$ (This is positive, meaning it is speeding up.)
2. $\fbox{$(\Delta \theta)=56.3\,\mbox{rad}=8.95\,\mbox{rev}$}$
3. $\fbox{$\alpha=-168\,\mbox{s}^{-2}$}$ (I represent the slowing down motion using the minus sign.)

For example, $\omega_0=800\,\mbox{RPM}=83.8\,\mbox{rad/s}=83.8\,\mbox{s}^{-1}$. There are $2\pi$ radians in one revolution or cycle.

Problem 1

1. $v_S'=28\,\mbox{m/s}$ (They ask for the speed. You should also be aware that this smaller puck moves off to the right.)
2. $\mbox{KE}_i=75\,\mbox{J}\qquad\mbox{and}\qquad\mbox{KE}_f=8\,\mbox{J}+67\,\mbox{J}=75\,\mbox{J}$. It is a perfectly elastic collision because $\mbox{KE}_i=\mbox{KE}_f$.
3. [make the drawing] The momenta are $p_S=3.57\,\mbox{kg}\,\mbox{m/s}$ to the right, $p_S'=1.19\,\mbox{kg}\,\mbox{m/s}$ to the left, and $p_L'=2.38\,\mbox{kg}\,\mbox{m/s}$ to the right.
4. [the answers are indicated in the previous part.] $v_S'=7\,\mbox{m/s}$ to the left.

Problem 1

1. [show this] Hint, unit of power is Watt=Joule/second, and unit of Joule=Nm.
2. [show this] Hint, use definition of work and $v=(\Delta x)/(\Delta t)$.
3. I get $\bar{F}_{\mbox{resistive}}=265\,\mbox{N}$.

Problem 2

1. I get $v'=-0.4\,\mbox{m/s}$ (the minus sign means it is moving to the left).
2. I get $F_{\mbox{ave}}=10\,\mbox{kN}$.
3. Hint, I get a total initial kinetic energy of $\mbox{KE}_i=7720\,\mbox{J}$, and a final value of $\mbox{KE}_f=21\,\mbox{J}$. The question asks for the individual kinetic energies.
4. Further question, what is the rate at which that kinetic energy was converted to other forms of energy? The "Power" in the collision? It happened in $\Delta t=0.10\,\mbox{s}$. I get $P=77\,\mbox{kW}$. The followup question is where did all that kinetic energy go?

Problem 3

Problem 1

1. $v=0.46\,\mbox{m/s}$
2. $x=+4.0\,\mbox{cm}$

Problem 2

1. Conservation of energy is the reason ... together with this being a projectile motion problem with no effects due to the air (like air resistance or buoyancy). As also shown in the next part, the speed of the water balloon at two elevations is related by $v_1^2 = v_0^2 + 2gh$ where $v_0$ is the slower speed at height $h$ above where the faster speed is $v_1$. It does not depend on the launch angle, which can be anywhere from straight up to straight down.
2. Follow the equations, where Event 0 is when it is just after it is launched and Event 1 is just before it hits the ground: \begin{align} KE_0 + PE_{g0} &= KE_1 + PE_{g1} \\[2ex] \tfrac12\,m\,v_0^2 + mgy_0 &= \tfrac12\,m\,v_1^2 + mgy_1 \\[2ex] v_0^2 + 2mgh &= v_1^2 + 2mg\cdot(0) \\[2ex] v_0 = \sqrt{v_1^2 - 2mgh} \end{align}

   I get $\fbox{$v_0=45\,\mbox{m/s}$}$.

Problem 1

1. 

2. The static friction force causes Manuel to move in a circle. (If there were suddenly no static friction, Manuel would move in a straight line until he reaches the perimeter of the carousel.)
3. I get $R=2.0\,\mbox{m}$.
4. With the same static coefficent of friction (which is why all the fuss about the shoes), that radius should not depend on the mass – within limits – of the person.

Problem 2

1. $N_{\mbox{top}}=(637)-(40)= 597\,\mbox{N}$ (straight up)
2. $N_{\mbox{bottom}}=(637)+(40)= 677\,\mbox{N}$ (straight up)
3. $N_{\mbox{halfway}}=\sqrt{(637)^2+(40)^2} = 638\,\mbox{N}$ (mostly up, but tipped $3.6^\circ$ toward the center)
4. The normal forces are not all equal. The normal force is computed using \[ \Sigma\vec{F}=\vec{N}+\vec{W}=m\vec{a} \] and with this problem, $\vec{W}$ a constant 637-N force pointing straight down and $\Sigma\vec{F}=m\vec{a}$ is the 40-N net force pointing toward the center of the Ferris wheel.

Problem 3

1. $\mbox{Work}_A=280\,\mbox{J}$
2. $\mbox{Work}_A=243\,\mbox{J}$
3. $\mbox{Work}_A=280\,\mbox{J}$
4. $\mbox{Work}_A=280\,\mbox{J}$
5. $\mbox{Work}_A=243\,\mbox{J}$
6. $\mbox{Work}_A=280\,\mbox{J}$

To remove the ambiguity in this problem, assume that the box is initially not moving.

Being pedantic, forces don't do work. The thing that exerts that force does the work, which is the amount by which the energy changes. If I lift a heavy box, it is I who gets tired, not the normal force I exert.

These problems are not intuitive unless you are a Newtonian thinker who correctly links forces to motion via N2L $(\Sigma\vec{F})=m\vec{a}$ and N3L $\vec{F}(\mbox{A on B})=-\vec{F}(\mbox{B on A})$.

Problem 1

These problems are not intuitive unless you are a Newtonian thinker who correctly links forces to motion via N2L $(\Sigma\vec{F})=m\vec{a}$ and N3L $\vec{F}(\mbox{A on B})=-\vec{F}(\mbox{B on A})$.

Problem 1

1. to the right, static friction
2. to the left, kinetic friction
3. the value of the static friction is zero. There is no slipping, so it is static friction, and the friction has no direction because it is zero. Note that it is not kinetic friction because the box is not sliding, even though it is moving relative to the road (as the van drives down the road).
4. to the West, static friction
5. to the West, static friction

Problem 2

You should also be able to figure out the following:

• $W=441\,\mbox{kN}$ (weight)
• $N=400\,\mbox{kN}$ (normal force)
• $F_k=240\,\mbox{kN}$ (kinetic friction)
• $a = 9.5\,\mbox{m/s}^2$ (acceleration magnitude as it slows down)

Details <SHOW>

Problem 3

Details <SHOW>

$ maxima

(%i1) sol:solve([T*cos(t)-muk*N=m*a, N+T*sin(t)-m*g=0],[N,T]);

                 g m cos(t) - a m sin(t)         g m muk + a m
(%o1)      [[N = -----------------------, T = -------------------]]
                   muk sin(t) + cos(t)        muk sin(t) + cos(t)


(%i2) float(subst([muk=0.60,m=1.3,t=55*%pi/180,g=9.8,a=2.4],sol));

(%o2)         [[N = 4.461321647756916, T = 10.10640016094792]]

These problems are not intuitive unless you are a Newtonian thinker who correctly links forces to motion via N2L $(\Sigma\vec{F})=m\vec{a}$ and N3L $\vec{F}(\mbox{A on B})=-\vec{F}(\mbox{B on A})$.

Problem 1

For the three-mass problem at hand, the solution I get is

1. $ a = 2.1\,\mbox{m/s}^2\qquad \mbox{(magnitude of acceleration of each mass)} $
2. $T_1 = 23.8\,\mbox{N}\qquad\mbox{(tension in string between A and B)}$
   $T_2 = 38.5\,\mbox{N}\qquad\mbox{(tension in string between B and C)}$
3. There are various ways to express this answer. Here is one: \begin{align} 19.6\,\mbox{N} = W_A \gt T_1 = 23.8\,\mbox{N} \\[2ex] 49.0\,\mbox{N} = W_C \lt T_2 = 38.5\,\mbox{N} \end{align} The intended point of this question is that the tension is not equal to the mass hanging on the string if that hanging mass is accelerating.

More Details <SHOW>

> solve([ T1 - mA*g = mA*a, T2 - T1 = mB*a, mC*g - T2 = mC*a], [a,T1,T2]);

                                  g (mA - mC)        g mA (mB + 2 mC)       mC g (2 mA + mB)
                         [[a = - ------------, T1 = ----------------, T2 = ----------------]]
                                 mA + mB + mC         mA + mB + mC           mA + mB + mC

(%i1) solve([ T1 - mA*g = mA*a, T2 - T1 = mB*a, mC*g - T2 = mC*a], [a,T1,T2]);

              g mA - g mC        g mA (2 mC + mB)       g mB mC + 2 g mA mC
(%o1) [[a = - ------------, T1 = ----------------, T2 = -------------------]]
              mC + mB + mA         mC + mB + mA            mC + mB + mA

Easy way to solve this problem <SHOW>

These problems are not intuitive unless you are a Newtonian thinker who correctly links forces to motion via N2L $(\Sigma\vec{F})=m\vec{a}$ and N3L $\vec{F}(\mbox{A on B})=-\vec{F}(\mbox{B on A})$.

Problem 1

Problem 2

1. Reading greater than Weight
2. Reading less than Weight
3. Reading equals the Weight
4. Reading equals the Weight
5. $a=1.2\,\mbox{m/s}^2$ Up
6. $a=1.4\,\mbox{m/s}^2$ Down

Problem 1

Remember that the units of \[ F = ma \qquad\qquad\mbox{are}\qquad\qquad \mbox{N}=\mbox{kg}\,\mbox{m/s}^2 \]

Problem 2

1. $(\Sigma F) = 8.1\times 10^6\,\mbox{N} $
2. The acceleration is $a_x=0.43\,\mbox{m/s}^2$, which is the x-component to the right; whle $a_y=0.00\,\mbox{m/s}^2$ is the y-component. \begin{align} F_x(\mbox{Train on Passenger}) &= +33\,\mbox{N} \qquad \mbox{(to the right, called static friction force)}\\[2ex] F_y(\mbox{Train on Passenger}) &= +755\,\mbox{N} \qquad \mbox{(straight up, called the normal force)}\\[2ex] \end{align}
3. Newton's third law can be used: \begin{align} F_x(\mbox{Passenger on Train}) &= -33\,\mbox{N} \qquad \mbox{(to the left, called static friction force)}\\[2ex] F_y(\mbox{Passenger on Train}) &= -755\,\mbox{N} \qquad \mbox{(straight down, called the normal force)}\\[2ex] \end{align}

Problem 1

Problem 2

1. I get \[ v_{0x}=19.6\,\mbox{m/s} \qquad \mbox{and}\qquad v_{0y}=12.5\,\mbox{m/s} \] This means \[ v_0 = 23.2\,\mbox{m/s} \qquad \mbox{and}\qquad \theta = 32.5^\circ \]
2. The x-motion and y-motion are independent.

Problem 3

1. I get for the initial velocity (at the hands): \[ v_{0x}=6.14\,\mbox{m/s} \qquad \mbox{and}\qquad v_{0y}=5.00\,\mbox{m/s} \] This means \[ v_0 = 7.92\,\mbox{m/s} \qquad \mbox{and}\qquad \theta_0 = 39.2^\circ \] which is moving upward and to the right, away from the hands.
2. I get for the final velocity (at the hoop): \[ v_{1x}=6.14\,\mbox{m/s} \qquad \mbox{and}\qquad v_{1y}=-1.86\,\mbox{m/s} \] This means \[ v_1 = 6.42\,\mbox{m/s} \qquad \mbox{and}\qquad \theta = -16.9^\circ \] which is moving downward and to the right, into the basket.

These are some answers I got:

Problem 1

1. $C=B=A$
2. $C=B=A$
3. $C>B>A$
4. $C>B>A$
5. (none)

Problem 2

The one launched at $30^\circ$ hit first and the time difference is $5.9\,\mbox{s}$.

You should be able to sketch the diagrams for each of these angles.

You should confirm the other data that I give above in the table.

Another thing you can do is figure out how the potato will move on Earth where $g=9.8\,\mbox{m/s}^2$ (while still disregarding effects due to the air, like wind resistance and buoyancy).

And one more thing you can do is to convert the $v_0=30\,\mbox{m/s}$ to MPH (miles/hour) to help develop an intuition for how fast that potato is moving.

These are some answers I got:

Problem 1

Problem 2

For more practice, you can figure out the initial (at B) and final (at A) velocities, \begin{align} v_{B} &= 2.35\,\mbox{m/s}\qquad \mbox{up the incline} \\[1ex] v_{A} &= 3.00\,\mbox{m/s}\qquad \mbox{down the incline} \end{align}

Problem 3

1. $300\,\mbox{m}$
2. $33^\circ$
3. Bob, $(\Delta t)=\left|(183)-(167)\right|=16\,\mbox{s}$
4. Displacements are equal to $330\,\mbox{m}$, but distances are not equal: Alice goes $330\,\mbox{m}$ and Bob goes $368\,\mbox{m}$.

I get the numerical answer of $s=3.02\,\mbox{m}$.

more <SHOW>

These are some final numerical answers I got for Problem 3:

2. The three non-zero accelerations are \[ a = \begin{cases} -5.09\,\mbox{m/s}^2 & 0.00\lt t\lt 3.00\,\mbox{s}\\[1ex] -1.39\,\mbox{m/s}^2 & 8.00\lt t\lt 10.00\,\mbox{s}\\[1ex] +9.65\,\mbox{m/s}^2 & 17.10\lt t\lt 17.39\,\mbox{s} \end{cases} \] This assumes that the final abrupt stop in those forty centimeters is taken into account. The acceleration from 3 to 8 seconds and from 10 to 17.1 seconds is zero.
3. $22.9\,\mbox{m}$
4. $17.39\,\mbox{s}$
5. $0.00\,\mbox{m/s}$
6. $2.63\,\mbox{m/s}=9.49\,\mbox{KPH}$